cf#257(Div.2)A.JzzhuandChildren_html/css_WEB-ITnose

A. Jzzhu and Children

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to n. The i-th child wants to get at least ai candies.

Jzzhu asks children to line up. Initially, the i-th child stands at the i-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:

1. Give m candies to the first child of the line.
2. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home.
3. Repeat the first two steps while the line is not empty.

Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?

Input

The first line contains two integers n,?m (1?≤?n?≤?100; 1?≤?m?≤?100). The second line contains n integers a1,?a2,?...,?an (1?≤?ai?≤?100).

Output

Output a single integer, representing the number of the last child.

Sample test(s)

input

5 21 3 1 4 2

output

input

6 41 1 2 2 3 3

output

Note

Let's consider the first sample.

Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home. Child 4 is the last one who goes home.

题意：要给n个人发糖果，每人发m颗；糖果有无限颗，但是每个人需要的糖果数不同，所以要求这n个人自觉排队，自热是从1到n的顺序；如果第i个人拿到的糖果数目大于m则回家，否则继续排队直到拿满。问最后一个走的人是谁？

解题思路：求出每个人要拿满至少需要排几次队，取最大的次数的人，如果有相同的则输出最后那个的编号！

#includeint main(){    int n,m,i,max=0,k;    int a[120];    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)    {        scanf("%d",&a[i]);        if(a[i]%m!=0)a[i]=(a[i]/m)+1;        else a[i]=a[i]/m;        if(a[i]>=max)        {            max=a[i];            k=i;        }    }    printf("%d\n",k);    return 0;}